ma 201: partial differential equations lecture - 7 · 2018-10-11 · lecture - 7 ma201(2018):pde....
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MA 201: Partial Differential EquationsLecture - 7
MA201(2018):PDE
Second Order Linear Equations
• Consider the second order linear equation in two independentvariables
(Auxx + Buxy + Cuyy + Dux + Euy + Fu)(x , y) = −G(x , y). (1)
• In operator notation
(T (u))(x , y) = −G(x , y) = f (x , y) (Say), (2)
withT (u) = Auxx + Buxy + Cuyy + Dux + Euy + Fu.
• Since T is linear (why?), we have
T (u1 + u2) = T (u1) + T (u2) & T (cu) = cT (u) ∀c ∈ R. (3)
Remark:
• Equation (2) is called homogeneous, if f = 0, otherwise it is callednon-homogeneous.
MA201(2018):PDE
The Principle of Superposition
Theorem
Suppose u1 solves linear PDE T (u) = f1 and u2 solves T (u) = f2, thenu = c1u1 + c2u2 solves T (u) = c1f1 + c2f2. In particular, if u1 and u2both solve the same homogeneous linear PDE T (u) = 0, so doesu = c1u1 + c2u2.
Remark:
• Any linear combination of solutions of linear homogeneous PDE isalso a solution.
• A solution u = u(x , y) to a homogeneous equation T (u) = 0 iscalled general solution if it contains two arbitrary functions.
• If u is a general solution to homogeneous equation T (u) = 0 and upis a particular solution to non-homogeneous equation T (w) = f ,then u + up is also a solution to the non-homogeneous equation andit is called general solution to T (w) = f .
MA201(2018):PDE
Linear Equations with Constant Coefficients
With the notations D = ∂/∂x and D ′ = ∂/∂y , a PDE with constantcoefficients can be written as
F (D,D ′)u = f . (4)
We further classify the PDE (4) into two main types:
• Reducible: Equation (4) is called reducible if it can be written asthe product of linear factors of the form aD + bD ′ + c , withconstants a, b, c . For example, the equation
uxx − uyy = 0.
In this case
F (D,D ′) = D2 − (D ′)2 = (D + D ′)(D − D ′).
• Irreducible: Equation (4) is called irreducible if it not reducible. Forexample when F (D,D ′) = D2 − D ′.
MA201(2018):PDE
Linear Equations with Constant Coefficients: Reducible Equation
An n-th order reducible PDE can be written as
F (D,D ′)u =(
n∏
r=1
(arD + brD′ + cr )
)
u = f . (5)
Theorem
If (arD + brD′ + cr ) is a factor of F (D,D ′), ar 6= 0, then
ur = exp
{
−crx
ar
}
φr (brx − ary)
is a solution of the equation F (D,D ′)u = 0. Here, φr is an arbitrary realvalued single variable function.
Theorem
If (brD′ + cr ) is a factor of F (D,D ′) and φr is an arbitrary real valued
single variable function, then
ur = exp
{
−cry
br
}
φr (brx)
is a solution of the equation F (D,D ′)u = 0.MA201(2018):PDE
Linear Equations with Constant Coefficients: Reducible Equation
Theorem
If (aD + bD ′ + c)m (m ≤ n, a 6= 0) is a factor of F (D,D ′) andφ1, φ2, . . . , φm are arbitrary real valued single variable functions, then
exp{
−cx
a
}
m∑
i=1
x i−1φi (bx − ay)
is a solution of the equation F (D,D ′)u = 0.
Theorem
If (bD ′ + c)m (m ≤ n) is a factor of F (D,D ′) and φ1, φ2, . . . , φm arereal valued single variable functions, then
exp{
−cy
b
}
m∑
i=1
x i−1φi (bx)
is a solution of the equation F (D,D ′)u = 0.
MA201(2018):PDE
Reducible Equations: Examples
Example
• General solution ofuxx − uyy = 0
is given byu = φ1(x − y) + φ2(x + y),
φ1 and φ2 are arbitrary real valued single variable functions.
• General solution of
∂4u
∂x4+∂4u
∂y4= 2
∂4u
∂x2∂y2
is given by
u = xφ1(x − y) + φ2(x − y) + xψ1(x + y) + ψ2(x + y).
MA201(2018):PDE
One Dimensional Wave Equation
MA201(2018):PDE
Vibrating string and the wave equation
• Consider a stretched string of length π with the ends fastened to the ends x = 0and x = π.
• Suppose that the string is set to vibrate by displacing it from its equilibriumposition.
• Let u(x , t) denote the transverse displacement at time t ≥ 0 of the point on thestring at position x . That is, we assume that each point of the string moves onlyin the vertical direction.
• In particular, u(x , 0) denotes the initial shape of the string and ut(x , 0) denotesthe initial velocity.
• That is, the string is set to vibrate by supplying an initial velocity ut(x , 0) to itfrom its equilibrium position u(x , 0) and then releasing it.
MA201(2018):PDE
Vibrating string and the wave equation (Contd.)• Under some physical assumptions, we arrive at the following equation known asthe one-dimensional wave equation:
utt(x , t) = c2uxx(x , t), (6)
which governs the entire process.
• Here c represents a physical quantity.
Note:
• The above equation is derived without considering any external force acting onthe string.
• Suppose that we do wish to include such an external force on the string (due toits weight or other impressed external forces (like gravity or pressure)), referredto as a load, given by
f (x , t) = vertical force per unit length at point x , at time t.
MA201(2018):PDE
Vibrating string and the wave equation (Contd.)• Then, the result will be the nonhomogeneous wave equation
utt − c2uxx = F (x , t), (7)
where F (x , t) = 1ρf (x , t), with ρ as the mass per unit length of the string.
Two special cases of (7) particularly generate interest:
• When the external force is due to the gravitational acceleration g only, theequation becomes
utt = c2uxx − g . (8)
• When the external force is due to the resistance of the medium (a stringvibrating in a fluid), the equation becomes
utt − c2uxx = −kut , k is a positive constant, (9)
known as damped wave equation.
MA201(2018):PDE
Wave EquationIt is to be noted that the use of string problem to demonstrate the waveequation is a matter of convenience. There are more applications inPhysics and Engineering.
For instance,u(x , t) = sin(x ± ct)
represents sinusoidal waves traveling with speed c in the positive andnegative directions, respectively, without change of shape.
Figure : Water Wave
MA201(2018):PDE
Wave Equation
Figure : Sound Wave
MA201(2018):PDE
Wave Equation (Contd.)
• Consider one-dimensional homogeneous wave equation
utt = c2uxx , (10)
• The solution is given by
u(x , t) = f (x − ct) + g(x + ct) (11)
where f and g are arbitrary real valued single variable functions.
• The physical interpretations of the functions f and g are quiteinteresting.
MA201(2018):PDE
Wave Equation (Contd.)To see this let us first consider the solution u = f (x − ct).
At t = 0, it defines the curve u = f (x) = u0(x)(say ), and after timet = t1, it defines the curve u = f (x − ct1) = u1(x)(say ).
But these curves u0 and u1 are identical in the sense that
u1(x) = f (x − ct1) = u0(X ),
that is, the origin is shifted by a distance ct1 towards right.
Figure : A Progressive Wave
MA201(2018):PDE
Wave equation (Contd.)For instance, consider
y (x , t) = sin(x − ct)
• A particle at the location (0, 0) at time t = 0 moves to location(ct1, 0) at time t = t1.
MA201(2018):PDE
Wave Equation (Contd.)Thus the entire configuration moves along the positive direction of thex-axis a distance of ct1 in time t1.
The velocity with which the wave is propagated is, therefore,
v =ct1
t1= c
Similarly the function g(x + ct) defines a wave progressing in thenegative direction of the x-axis with constant velocity c.
The total solution is, therefore, the algebraic sum of these two travelingwaves.
Solution (11) is a very convenient representation for progressive waveswhich travel large distances through a uniform medium.
MA201(2018):PDE
The D’Alembert solution of the wave equationLet us consider the following initial value problem
utt − c2uxx = 0, −∞ < x <∞, (12)
Displacement: u(x , 0) = φ(x), (13)
Velocity: ut(x , 0) = ψ(x), (14)
that is, we consider the vibration of a thin string of infinite length withsome initial displacement and initial velocity.
From solution (11), we find that
f (x) + g(x) = φ(x), (15)
−cf ′(x) + c g ′(x) = ψ(x), (16)
for all values of x .
MA201(2018):PDE
The D’Alembert solution of the wave equation (Contd.)Integrating the second equation with respect to x
−f (x) + g(x) =1
c
∫ x
x0
ψ(τ)dτ + A. (17)
where A is an integration constant and τ is a dummy variable.
From (15) and (17)
f (x) =1
2
[
φ(x)−1
c
∫ x
x0
ψ(τ)dτ
]
− A/2, (18)
g(x) =1
2
[
φ(x) +1
c
∫ x
x0
ψ(τ)dτ
]
+ A/2, (19)
Substituting these expressions into (11)
u(x , t) =1
2[φ(x − ct) + φ(x + ct)] +
1
2c
∫ x+ct
x−ct
ψ(τ)dτ. (20)
This is D’Alembert’s (1717-1783) solution of one-dimensional wave
equation, which he actually derived around 1746.
MA201(2018):PDE
Non-Homogeneous Wave Equation: Duhamel’s PrincipleFind u(x , t) such that
utt(x , t)− c2uxx = F (x , t), (x , t) ∈ (−∞,∞)× (0,∞) (21)
with ICs u(x , 0) = 0, ut(x , 0) = 0 ∀x ∈ (−∞,∞). (22)
Then the solution u is given by
u(x , t) =
∫ t
0
v(x , t − τ, τ)dτ. (23)
What is v?
MA201(2018):PDE
Duhamel’s Principle (Contd.)Here, v is the solution of the problem
vtt − αvxx = 0, (x , t) ∈ (0, L)× (0,∞) (24)
with ICs v(x , 0) = 0, vt(x , 0) = f (x , s), s > 0, (25)
for some real parameter s > 0.
Note that the solution v of the above problem depends on x , t andparameter s.
Thus, v = v (x , t, s). Accordingly, we can modify the IC’s as
v(x , 0, s) = 0 = φ (say), vt(x , 0, s) = f (x , s) = ψ(x) (say), s > 0.
Due to D’Alembert’s, v = v (x , t, s) is given by
v(x , t, s) =1
2c
∫ x+ct
x−ct
ψ(τ)dτ =1
2c
∫ x+ct
x−ct
f (τ, s)dτ.
MA201(2018):PDE
Leibniz Rule: Differentiation under the sign of integration
d
dt
(
∫ b(t)
a(t)
G (x , t)dx)
=
∫ b(t)
a(t)
Gtdx + G (b(t), t)b′(t)
−G (a(t), t)a′(t). (26)
Use Leibniz rule on
u(x , t) =
∫ t
0
v (x , t − τ, τ)dτ
to have
ut =
∫ t
0
vt(x , t − τ, τ)dτ + v(x , t − t, t)×dt
dt− v(x , t − 0, 0)× 0
=
∫ t
0
vt(x , t − τ, τ)dτ + v(x , 0, t) =
∫ t
0
vt(x , t − τ, τ)dτ.
Similarly
utt(x , t) =
∫ t
0
vtt(x , t − τ, τ)dτ + vt(x , 0, t) =
∫ t
0
vtt(x , t − τ, τ)dτ + f (x , t)
c2uxx (x , t) = c2∫ t
0
vxx(x , t − τ, τ)dτ so that utt − c2uxx = f (x , t).
MA201(2018):PDE
Completely Non-Homogeneous Wave Equation: Duhamel’s
PrincipleFind u(x , t) such that
utt(x , t)− c2uxx = F (x , t), (x , t) ∈ (−∞,∞)× (0,∞) (27)
with ICs u(x , 0) = φ(x), ut(x , 0) = ψ(x) ∀x ∈ (−∞,∞). (28)
Try to break the problem as (ρ, θ) problems as: Find θ such that
θtt(x , t)− c2θxx = F (x , t), (x , t) ∈ (−∞,∞)× (0,∞) (29)
with IC’s θ(x , 0) = 0, θt(x , 0) = 0 ∀x ∈ (−∞,∞). (30)
Find ρ such that
ρtt(x , t)− c2ρxx = 0, (x , t) ∈ (−∞,∞)× (0,∞) (31)
with IC’s ρ(x , 0) = φ(x), ρt(x , 0) = ψ(x) ∀x ∈ (−∞,∞). (32)
Principle of Super Position tells that ρ+ θ solves the completelynon-homogeneous problem.
MA201(2018):PDE